Give me a concrete example of an overload you want to write but wouldn't be able to if `TypeGaurd` is considered a subtype of bool by type checkers. (again regardless of what type checkers think typeguard variables will be bools so `isinstance(guard, bool)` is going to be True and `isinstance(guard, TypeGuard)` is going to raise TypeError).
On Wed, Mar 17, 2021 at 5:41 PM Jake Bailey via Typing-sig < firstname.lastname@example.org> wrote:
Except then no existing function (like filter) could ever get a new overload with TypeGuard, because the compatibility would make it illegal, reducing it's usefulness. _______________________________________________ Typing-sig mailing list -- email@example.com To unsubscribe send an email to firstname.lastname@example.org https://mail.python.org/mailman3/lists/typing-sig.python.org/ Member address: email@example.com